Skip to main content

Full text of "Treatise On Analysis Vol-Ii"

See other formats


14    LOCALLY CONVEX SPACES       65

exists a continuous mapping gmn of X into C, with support contained in
Ow+1 (and therefore compact) and agreeing with hmn on On. Consequently, for
each e > 0, each/e ^c(X) and each integer , there exists an integer m such
that pn(f gmn) g . Since the sequence of seminorms (/?) is increasing, this
proves the result, having regard to (12.14.3.1) and the definition of the topo-
logy on #CPQ.

Remark

(12.14.6.3) The proof shows also that if Supp(/) c Un, then/is the limit
of a sequence of functions gmn with supports contained in Un+1.

(12.14.7)    Let (/?a)a 61 be a family of seminorms on a vector space E. For each
finite subset H of I, the function j7H = sup pa is a seminorm on E (12.14.1).

aeH

Since the set W((af); r) defined in (12.14.3.1) is the set of all x e E such that
/?H(JC) < r (where H is the Set of the af), it follows that the family of seminorms
pH, where H runs through all finite subsets of I, is equivalent to the family
(pa). A set F of seminorms on E is said to be directed if for each pair of semi-
norms p', p" belonging to F, there exists p e T such that p g: sup(p', p"). It
follows that the topology of a locally convex space can always be defined by a
directed set of seminorms.

Let E be a locally convex space whose topology is defined by a set F of
seminorms. If F is a vector subspace of E, it is clear that the induced topology
on F is defined by the restrictions to F of the seminorms belonging to F. The
corresponding result for quotient spaces is the following:

(12.14.8)    Let E be a vector space, F a vector subspace o/E, and n:E-+ E/F
the canonical homomorphism.

(i)   Let p be a seminorm on E. For each x e E/F let

(12.14.8.1)                              p(*) = inf p(z).

Then p is a semi-norm on E/F.

(ii) If F is a directed set of seminorms and ifE is endowed with the topology
defined by F, then the quotient topology on E/F (12.11) is defined by the set of
seminorms /), where p e F.

(i)   We have

i = inf p(Az)  |A| inf p(z) ==

so that it remains to prove the triangle inequality. If x, y are elements of E/F there exists a dense